Suppose that two compact topological manifolds with boundary have homeomorphic interiors. Can we conclude that the two manifolds are homeomorphic? What happens in the smooth category?
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2 Answers
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Gjergji Zaimi's answer gives a strong positive conclusion: the product of the boundaries with $\mathbb{R}$ are necessarily homeomorphic. I just want to add a couple of explicit examples illustrating that the boundaries themselves need not be homeomorphic.
The first example is given by Milnor in his article "Two complexes which are homeomorphic but combinatorially distinct", partially stated there as theorem 3. Consider the 3-dimensional lens spaces $L_1=L(7,1)$ and $L_2=L(7,2)$. In that article, Milnor proves (section 2) that for any $n>1$, the interior of $L_1\times D^{2n+1}$ is diffeomorphic to the interior of $L_2\times D^{2n+1}$. However, Milnor also proves (at the end of section 4) that their boundaries, $L_1\times S^{2n}$ and $L_2\times S^{2n}$, have different Reidemeister torsions. Consequently, the boundaries cannot be homeomorphic, by the topological invariance of Reidemeister torsion.
To obtain other examples, we can apply the result stated by Oscar Randall-Williams' in his comment to Gjergji Zaimi's answer. It suffices to find $h$-cobordant closed manifolds $M$, $N$ of dimension greater than 3 which are not homeomorphic: then $M\times I$, $N\times I$ will have diffeomorphic interiors by Oscar's argument, but their boundaries are not homeomorphic. As Milnor proves in the article cited above, we can take $M=L_1\times S^{2n}$ and $N=L_2\times S^{2n}$ for $n>0$. I wrote a more detailed discussion concerning examples of this type in an answer to a related mathoverflow question.
answered Mar 30, 2013 at 20:58
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$\begingroup$ Thank you Ricardo, this is a very interesting answer, I didn't know this result of Milnor. $\endgroup$ Commented Mar 31, 2013 at 7:14
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$\begingroup$ @Daniele: You are most welcome. I am glad I could help. $\endgroup$ Commented Mar 31, 2013 at 10:25
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In "Uniqueness of the Open Cone Neighborhood" by K.W. Kwun, Proc. Amer. Math. Soc. 15 (1964) pp.476-479, it is shown that if two manifolds have homeomorphic interiors then their boundaries have homeomorphic suspensions, or in other words the boundaries are h-cobordant. I don't know if one can conclude anything more in this generality.
answered Apr 25, 2012 at 9:28
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8$\begingroup$ I don't think one can say any more generally: If $W : M \leadsto N$ is an $h$-cobordism then $S^1 \times W$ is an $s$-cobordism (crossing with $S^1$ kills Whitehead torsion), and taking covering space corresponding to the subgroup $\pi_1(W) \leq \pi_1(S^1 \times W)$ shows that $\mathbb{R} \times M \cong \mathbb{R} \times N$. In particular, $[0,1] \times M$ and $[0,1] \times N$ have isomorphic interiors. $\endgroup$ Commented Apr 25, 2012 at 13:36
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$\begingroup$ @Oscar. The conclusion that the two boundaries times $R$ are homeomorphic is the theorem of Kwun paper cited by Gjergji. $\endgroup$ Commented Apr 25, 2012 at 14:49